kitchen table math, the sequel: help desk, part 3 slope again

Wednesday, June 27, 2007

help desk, part 3 slope again

Steve H left this demonstration (proof?) of why, when calculating slope, it's OK to use either point as "Point 1":

(y1-y2)/(x1-x2)

= -(y2-y1)/[-(x2-x1)]

= -1*(y2-y1)/[-1*(x2-x1)]

= -1/-1 * (y2-y1)/(x2-x1)

= (y2-y1)/(x2-x1)


Here's my question.

What are the justifications for each step?

(y1-y2)/(x1-x2)

= -(y2-y1)/[-(x2-x1)] mutiplicative identity ??

= -1*(y2-y1)/[-1*(x2-x1)] mutiplicative identity again??

= -1/-1 * (y2-y1)/(x2-x1) commutative property of multiplication??

= (y2-y1)/(x2-x1)


I'm definitely ready for something more formal...although I'm not sure where I'm going to find it.

Think I'll check Dolciani's & Foerster's books.


update from Steve:

I think schools should spend much more time on these identities. They should show how they are used in all sorts of interesting ways, forwards and backwards. I don't think I really learned algebra until this happened. [Catherine here: I agree absolutely. I'm really feeling a need for this, and it's not really something I can "provide" for myself - at least, not without a HUGE amount of effort.]


(y1-y2)/(x1-x2)

= -(y2-y1)/[-(x2-x1)]

This is the Distributive Law, in reverse, if you will.

[Catherine: I didn't see this! Now I do!]


= -1*(y2-y1)/[-1*(x2-x1)]

I suppose you could call this the Multiplicative Identity. I mentioned before that the sign "belongs" to the term that comes after it. I always like to think of a sign as a +1 or a -1. [Catherine: me, too]

= -1/-1 * (y2-y1)/(x2-x1)

I suppose you could call this the Commutative Property of Multiplication law. In other words, it doesn't matter which way you multiply fators. That's why I liked to have students circle the factors in a rational term. Then they know that they can move the factors to any position.


= (y2-y1)/(x2-x1)

I got rid of the -1/-1 using the Multiplicative Inverse law.


It's interesting that in the list of basic identities I found online that there wasn't an identity for

a/1 = a

I suppose this would come from the Definition of Division

3 comments:

SteveH said...

I think schools should spend much more time on these identities. They should show how they are used in all sorts of interesting ways, forwards and backwards. I don't think I really learned algebra until this happened.


(y1-y2)/(x1-x2)

= -(y2-y1)/[-(x2-x1)]

This is the Distributive Law, in reverse, if you will.


= -1*(y2-y1)/[-1*(x2-x1)]

I suppose you could call this the Multiplicative Identity. I mentioned before that the sign "belongs" to the term that comes after it. I always like to think of a sign as a +1 or a -1.

= -1/-1 * (y2-y1)/(x2-x1)

I suppose you could call this the "Commutative of Multiplication law. In other words, it doesn't matter which way you multiply fators. That's why I liked to have students circle the factors in a rational term. Then they know that they can move the factors to any position.


= (y2-y1)/(x2-x1)

I got rid of the -1/-1 using the Multiplicative Inverse law.


It's interesting that in the list of basic identities I found online that there wasn't an identity for

a/1 = a

I suppose this would come from the Definition of Division

Catherine Johnson said...

I think schools should spend much more time on these identities. They should show how they are used in all sorts of interesting ways, forwards and backwards. I don't think I really learned algebra until this happened.

I agree absolutely.

It's a big gap in my own knowledge.

Adrian said...

Actually, I think that (-1)*x equals the additive inverse of x usually requires a special proof -- something like x+(-1)x=[1+(-1)}x=0*x=0 ==> (-1)x=-x (which inturn is based on the fact that 0*x=0 for all x). Of course, it really depends on just how you develop it axiomatically. I have seen a few slightly different renditions of this development, mostly in abstract algebra texts, but also in new math texts from the 60s. (Well, actually, only in Frank Allen's text, but they might have it in an old 1960s copy of Dolciani.

I think you are right about why a/1=a -- it depends on what and when the definition of "/" is given. In an abstract algebra course, I imagine that such things as the "/" notation are just not worried about (either being omitted or just used without talking much about it). The issue of a/1=a would be whether the multiplicative inverse of the multiplicative identity is the multiplicative identity. The main theorem that you will use for that is that the inverse is unique and then it immediately follows that 1 inverse is 1 since 1*1=1.